Maths Methods 1/2 Help!

[stripe]

Hi everyone,
I am currently doing Maths Methods 1/2 and am having some trouble with a couple of questions.  Can anyone help please?
the first one is :

A cubic function has rule P(x) = ax³ + bx².  Its graph passes through points (1,1) and (-3,-63).  Show that a=2 and b= -1.

I'm not sure what I'm doing wrong but I just can't get it.
If you could help that would be great.
Thank you.

[DisaFear]

Substitute (1,1) into the formula
You get 1=a+b
Let this be equation 1

Substitute (-3, -63) into the formula
You get -63=-27a+9b
Let this be equation 2

Re-arrange equation 1 as a=1-b

Now substitute equation 1 into equation 2, and solve!

-63=-27+27b + 9b (substituted equation 1 into equation 2)
-36=36b
b=-1

a=1-b  (equation 1, rearranged)
a=2

[stripe]

Thank you so much!

[stripe]

What about this question?

Consider the cubic function y = (x+2)(x² + 3x + 7)
Show that 0 = (x+2)(x² + 3x + 7) has only one real solution.

I know I have to use the discriminant but how?

Thanks for your help

[Nobby]

What about this question?

Consider the cubic function y = (x+2)(x² + 3x + 7)
Show that 0 = (x+2)(x² + 3x + 7) has only one real solution.

I know I have to use the discriminant but how?

Thanks for your help

=> (x² + 3x + 7)
Find the discriminant of that. It should be negative. Therefore, there are no solutions for x² + 3x + 7 = 0, and thus the only solution for the cubic function is x = -2.

[Greatness]

x+2=0 or x^2+3x+7=0, use the discriminant with the 2nd eqn. So Discriminant=b^2-4ac = 9 - 4*1*7=9-28 which is less than 0. Therefore there is only one real solution which is x=-2

[stripe]

Thanks very much!

[stripe]

What about this one?

If y = (x+2)(x²+3kx+7), where k is a constant, find the values of k such that 0 = (x+2)(x²+3kx+7), has
2 real solutions;
3 real solutions

Thanks for your help!

[DisaFear]

The (x+2) will always give one solution
Hence, we need to find when the discriminant of the other part of the equation, the one with the k, that will give us 1 and 2 solutions

The discriminant is (3k)^2 -28
To get one solution, let it equal zero
k=sqrt(28)/3

To get two solutions, let discriminant > 0
k>sqrt(28)/3 and k<-sqrt(28)/3

I'm never too sure with my answers tell me if that's the answer

[stripe]

Thanks very much.
I got that k > sqrt(28)/3, but why would it also be < sqrt(28)/3 ?
Thanks again for all your help

[DisaFear]

Thanks very much.
I got that k > sqrt(28)/3, but why would it also be < sqrt(28)/3 ?
Thanks again for all your help

Yea, sorry, its k<-sqrt(28)/3, which is obtained by square rooting the right side (Thanks to ~My♥Little♥Pony~)

[stripe]

Oh, I understand, because the solution can be less than zero if the discriminant >0?

What about translation?  If I wanted to move the graph of y = 0.1x³ - 1.2x² + 4.1x + 0.8 by .5 down the y axis the answer would be 0.1x³ - 1.2x² + 4.1x + 0.3.  Is that right?

And my final question, if I have a few points on a graph (0, 0.3) (2,4) (5, 2. and (7,4) how do I find the equation?

Thank you so much for all your help

[DisaFear]

What about translation?  If I wanted to move the graph of y = 0.1x³ - 1.2x² + 4.1x + 0.8 by .5 down the y axis the answer would be 0.1x³ - 1.2x² + 4.1x + 0.3.  Is that right?

Yep! The y-intercept depends on that value without the x

And my final question, if I have a few points on a graph (0, 0.3) (2,4) (5, 2. and (7,4) how do I find the equation?

I assume that's a cubic graph?
Substitute each pair of values into the general cubic formula
You'll end up with 4 equations, each one has one pair subbed in, and has a,b,c,d
Then do a massive simultaneous solution to find the a,b,c,d (the coefficients of x and the y intercept)
I think

[dinosaur93]

And my final question, if I have a few points on a graph (0, 0.3) (2,4) (5, 2. and (7,4) how do I find the equation?

When doing simultaneous equation, you have 3 ways of finding the equation with 3 given points..

Firstly, BY HAND (but I think this is unlikely to be given as a question in a tech free test)

given 3 points, you assume that its a cubic equation hence taking the form of:

substitute the 3 points into the equation therefore generating 4 equations to be used in simultaneous equation.

Secondly, BY CAS  (1) - Main 2D Keyboard

then plot in the 4 equations into the CAS to generate the answer a, b, c, d.


Lastly, BY CAS (2) EASIEST POSSIBLE WAY -  Menu Stats (Plot in the points) Calc Cubic Reg OK

     

A cubic function has rule P(x) = ax³ + bx².  Its graph passes through points (1,1) and (-3,-63).  Show that a=2 and b= -1.

substitute (1,1) 1st equation
substitute (-3, -63) 2nd equation


1st equation



________________




hence sub in 2 in the 1st equation to get b

1 = 2 + b




hope this helps~

[stripe]

Thank you everyone for helping me. 
I really appreciate it.